Problem 32 - Discrete Mathematics with Applications

Question

Is \(\left\langle S, \cup, \cap,^{\prime}, \emptyset, U\right\rangle\) a boolean algebra for each subset \(S\) of \(P(U),\) where \(U=\\{a, b, c\\} ?\) $$\\{\emptyset, | a\\},\\{b\\},\\{a, b\\}, U\\}$$

Step-by-Step Solution

Answer

The set system \(\left\langle S, \cup, \cap,^{\prime}, \emptyset, U\right\rangle\) with \(S = \{ \emptyset, \{a\}, \{b\}, \{a, b\}, U\}\) and \(U = \{a, b, c\}\) is not a boolean algebra, since it does not satisfy the complementation axiom.

Step 1: Check Commutativity

We will check if the commutative laws hold for all pairs of sets in \(S\). Note that since the commutative laws state equalities, it's enough to verify for unordered pairs of distinct elements, as the others follow trivially. \(\{a\} \cup \{b\} = \{a, b\} = \{b\} \cup \{a\}\), which confirms commutativity for the union operation. \(\{a\} \cap \{b\} = \emptyset = \{b\} \cap \{a\}\), which confirms commutativity for the intersection operation. Commutativity axiom holds.

Step 2: Check Associativity

For associativity, we need to check if it holds for all triplets of sets in \(S\). - \((\{a\} \cup \{a\}) \cup \{b\} = \{a\} \cup \{b\} = \{a, b\} = \{a\} \cup (\{a\} \cup \{b\})\) - \((\{b\} \cap \{a\}) \cap \{b\} = \emptyset \cap \{b\} = \emptyset = \{b\} \cap (\{a\} \cap \{b\})\) Since S has only four non-trivial sets, the other cases follow directly. Associativity axiom holds.

Step 3: Check Identity

We will check if the identity laws hold for all the sets in S: - \(\emptyset \cup \emptyset = \emptyset\) - \(\emptyset \cap U = \emptyset\) - \(\{a\} \cup \emptyset = \{a\} \cap U = \{a\}\) - \(\{b\} \cup \emptyset = \{b\} \cap U = \{b\}\) - \(\{a, b\} \cup \emptyset = \{a, b\} \cap U = \{a, b\}\) Identity axiom holds.

Step 4: Check Complementation

It's important to note that we don't have explicit complements in the given set S. Since we don't have a proper complement for all sets in S, the complementation axiom does not hold.

Step 5: Conclusion

Since the given set system \(\left\langle S, \cup, \cap,^{\prime}, \emptyset, U\right\rangle\) does not satisfy the complementation axiom, it cannot be considered a boolean algebra.

Key Concepts

Commutative Law

The Commutative Law is a principle that applies to both union and intersection in set theory, much like addition and multiplication in arithmetic. It suggests that the order of the elements does not affect the outcome.
For example, if you have two sets, say \(A\) and \(B\), the union and intersection can be swapped:

• \(A \cup B = B \cup A\)
• \(A \cap B = B \cap A\)

In the given exercise, we checked these properties for the sets \(\{a\}\) and \(\{b\}\), confirming that the operation is commutative. This essential property ensures flexibility in how operations are performed, without changing results.

Associative Law

The Associative Law deals with how the grouping of sets affects the outcome of union and intersection operations. It allows you to group elements in different ways without changing the result.
For instance, given three sets \(A\), \(B\), and \(C\), both of the following equations hold:

• \((A \cup B) \cup C = A \cup (B \cup C)\)
• \((A \cap B) \cap C = A \cap (B \cap C)\)

This property helps in simplifying the expression of complex set operations by allowing rearrangement and grouping as required. In our exercise, checking associativity was done by confirming that different groupings of operations yield the same result, such as with \(\{a\}, \{b\}, \{a, b\}\). Association gives structure to boolean operations.

Identity Law

The Identity Law ensures that there are neutrality elements within set operations.
The neutral element for union is the empty set \(\emptyset\), while for intersection, it's the universal set \(U\). Here's how they work:

• \(A \cup \emptyset = A\)
• \(A \cap U = A\)

The identity element does not change the set it is combined with. For example, if you take any set \(\{a\}\) or \(\{b\}\) and apply these identity laws:

• \(\{a\} \cup \emptyset = \{a\}\)
• \(\{b\} \cap U = \{b\}\)

This characteristic forms the core idea of maintaining the original data in various operations, providing consistency within boolean algebra.

Complementation

Complementation refers to the idea of finding the opposite or complement of a set.
In a complete boolean algebra, each set \(A\) has a complement \(A'\), such that:

• \(A \cup A' = U\)
• \(A \cap A' = \emptyset\)

It involves the use of the universal set \(U\) to ensure that every element has a counterpart. However, in the given exercise, not all sets had complements represented, leading to the failure of the complementation property. A lack of proper complementation means we can't call the system a boolean algebra.

Set Theory

Set Theory forms the groundwork from which boolean algebra principles are drawn. It deals with collections of objects, known as sets, and operations like union, intersection, and complementation.
This foundational concept is crucial for various fields in mathematics and computer science. When applied to boolean algebra:

• Union \( \cup \) brings together elements from different sets.
• Intersection \( \cap \) finds common elements.
• Complementation flips elements' inclusion within the universal set \(U\).

Set theory enables logical structuring and reasoning with finite and infinite collections, proving indispensable whenever we formalize objects and rules governing them, just as shown in our discussed exercise. Understanding set theory aids in unraveling more complex logic operations.